What is the vertices 9x^2+36x+9y^2+72y+155=0 ?

The vertices 9x^2+36x+9y^2+72y+155=0 is (-2,-7/3),(-2,-17/3)

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vertices 9x^2+36x+9y^2+72y+155=0

vertices

9 x^{2} + 36 x + 9 y^{2} + 72 y + 155 = 0

(- 2, - \frac{7}{3}), (- 2, - \frac{17}{3})

Solution steps

(h, k + b), (h, k - b)

Calculate ellipse properties

9 x^{2} + 36 x + 9 y^{2} + 72 y + 155 = 0 :

Ellipse with center

(h, k) = (- 2, - 4), b = \frac{5}{3}, a = \frac{5}{3}

(- 2, - 4 + \frac{5}{3}), (- 2, - 4 - \frac{5}{3})

Refine

(- 2, - \frac{7}{3}), (- 2, - \frac{17}{3})

\frac{x ^{2}}{2 ^{2}} + \frac{y ^{2}}{3 ^{2}} = 1

x^{2} + 4 x + 4 y^{2} = 0

\frac{( x + 1 ) ^{2}}{36} + \frac{( y + 1 ) ^{2}}{64} = 1

4 x^{2} + y^{2} - 48 x - 4 y + 48 = 0

\frac{( x - 1 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{5} = 1

What is the vertices 9x^2+36x+9y^2+72y+155=0 ?
The vertices 9x^2+36x+9y^2+72y+155=0 is (-2,-7/3),(-2,-17/3)